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How to write the slope intercept form of the line $ 3x - 2y = - 16 $ ?

Answer
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Hint: In this question we have to write the slope intercept form of the line $ 3x - 2y = - 16 $ . We know that the equation in the slope intercept will be in the form like $ y = mx + b $ . So we will convert the given equation into this form i.e., we will solve for y.

Complete step-by-step answer:
Here, we have to write the slope intercept form of the line $ 3x - 2y = - 16 $ .
The equation in slope intercept form will be like $ y = mx + b $ , where $ m $ is the slope of the line and $ b $ is the $ y $ -intercept of the line, or the $ y $ -coordinate if the point at which the line crosses the $ y $ -axis.
To write the line $ 3x - 2y = - 16 $ in slope intercept form, we have to solve for $ y $ .
Therefore, $ - 2y = - 16 - 3x $
\[2y = 3x + 16\]
Then, \[y = \dfrac{{3x + 16}}{2}\]
 $ y = \dfrac{3}{2}x + \dfrac{{16}}{2} $
Hence, $ y = \dfrac{3}{2}x + 8 $
Where $ m = \dfrac{3}{2} $ (slope)
And, $ b = 8 $ ( $ y $ -intercept)
Therefore, the slope intercept form of the line $ 3x - 2y = - 16 $ is $ \dfrac{3}{2}x + 8 $ .
So, the correct answer is “ $ y = \dfrac{3}{2}x + 8 $ ”.

Note: In this question it is important to note here that $ y = mx + b $ is the form called the slope-intercept form of the equation of the line. It is the most popular form of straight line. Many find this as useful because of its simplicity. One can easily describe the characteristics of the straight line even without seeing its graph because the slope and $ y $ -intercept can easily be defined or read off from this form. The slope $ m $ measures how steep the line is with respect to the horizontal. Let us consider two points $ \left( {{x_1},{y_1}} \right) $ and $ \left( {{x_2},{y_2}} \right) $ found in the line, the slope can be written as, $ m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $ . The $ y $ -intercept $ b $ is the point where the line crosses the $ y $ -axis.